NonLinear Regression
Gabriel Santos / Anjal Hussan
2023-04-18
Dataset:
Cars and salespersons
we would like to analyze the relationship between how many total cars have been sold by each sales person and how many weeks each salesperson has worked to sell this cars.
Libraries
library(tidyverse)
library(ggplot2)
library(dplyr)SalesCars <- read_csv('https://raw.githubusercontent.com/GabrielSantos33/Non-linear-regression/main/CarsSoldWeek.csv', show_col_types = FALSE)Data Set
SalesCars| J_period | Cars_sold |
|---|---|
| 168 | 272 |
| 428 | 300 |
| 296 | 311 |
| 392 | 365 |
| 80 | 167 |
| 56 | 149 |
| 352 | 366 |
| 444 | 310 |
| 168 | 192 |
| 200 | 229 |
| 4 | 88 |
| 52 | 118 |
| 20 | 62 |
| 228 | 319 |
| 72 | 193 |
Plotting the data
ggplot(SalesCars, aes(x = J_period, y = Cars_sold)) +
geom_point() +
labs(title = "Scatter Plot",
x = "Job period",
y = "Total cars sold")
Does it look linear? Does it look non linear?, both?
What model best represents this data?
Does this linear regression fits the data quite well?
Linear model
ggplot(SalesCars, aes(x = J_period, y = Cars_sold)) +
geom_point() +
geom_smooth(method = "lm", formula = "y ~ x",se = FALSE) +
labs(title = "Scatter Plot Example",
x = "Job period",
y = "Total cars sold")
Fit linear regression model
lm_model <- lm(Cars_sold ~ J_period, data = SalesCars)Coefficients
coeficient <- coef(lm_model)Extract R-squared value
r_squared <- summary(lm_model)$r.squaredPredict y-values
predicted_y <- predict(lm_model)Adding R-squared value and predicted y-values to dataframe
SalesCars$r_squared <- r_squared
SalesCars$predicted_y <- predicted_yDisplay R-squared value and predicted y-values
cat("R-squared value: ", round(r_squared, 2), "\n")## R-squared value: 0.8cat("Predicted Y-values: ", paste(round(predicted_y, 2), collapse = ", "))## Predicted Y-values: 212.32, 363.71, 286.85, 342.75, 161.08, 147.1, 319.46, 373.03, 212.32, 230.95, 116.83, 144.77, 126.14, 247.26, 156.42Extract coefficients of the linear equation
intercept <- coef(lm_model)[1]
slope <- coef(lm_model)[2]
# Display the linear equation
cat("Linear equation: y =", round(intercept, 2), "+", round(slope, 2), "* x")## Linear equation: y = 114.5 + 0.58 * xFor each week a person has add a bit more than a half car per week
For every week the sells is expected to increase by 0.58 per week
NOTE: for rare cases where setting the intercept to ZERO makes sense. but for this case we will not set it this up.
Linear regression Output
print(summary(lm_model))##
## Call:
## lm(formula = Cars_sold ~ J_period, data = SalesCars)
##
## Residuals:
## Min 1Q Median 3Q Max
## -64.142 -27.800 1.896 30.364 71.743
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 114.49632 19.78813 5.786 6.32e-05 ***
## J_period 0.58228 0.08026 7.255 6.41e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 45.94 on 13 degrees of freedom
## Multiple R-squared: 0.8019, Adjusted R-squared: 0.7867
## F-statistic: 52.63 on 1 and 13 DF, p-value: 6.412e-06Linear equation: y = 114.5 + 0.58 * x Multiple R = 0.8955 very high Adjusted R-squared: 0.7867 R-squared: 0.8019 this model explain the 80% of the variance standard error: 45.94 how well the observation fit around the regression line
ANOVA results
anova_results <- anova(lm_model)print("ANOVA Results:")## [1] "ANOVA Results:"print(anova_results)## Analysis of Variance Table
##
## Response: Cars_sold
## Df Sum Sq Mean Sq F value Pr(>F)
## J_period 1 111097 111097 52.633 6.412e-06 ***
## Residuals 13 27440 2111
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Regression = mean square of 111097 Residuals = 27440 total sum of squares = 138537 ——— Mean Sq = 111097 Mean Sq error or residual = 2111 = low relative to the total sum of squares F Statistics = 52.633 very high you can se that our significance value is very small- the model is statically significant P-value = 6.32e-05 Jperiod = 6.41e-06
Residuals
Let’s see the residuals
residuals <- residuals(lm_model)
# Create a residuals plot
plot(residuals, main = "Residuals Plot", xlab = "Predicted Cars_sold", ylab = "Residuals")
#x = J_period, y = Cars_sold)Transforming the exploratory valiables
Simply by transforming the exploratory valiables does not make a non-linear model
ggplot(SalesCars, aes(x = J_period, y = Cars_sold)) +
geom_point() +
geom_smooth(method = "lm", formula = "y ~ log(x)",se = FALSE) +
labs(title = "Non linear Scatter Plot Example - transformation",
x = "Job period",
y = "Total cars sold")
Exponetial model
Lets use an exponetial model not a good model
ggplot(SalesCars, aes(x = J_period, y = Cars_sold)) +
geom_point() +
geom_smooth(method = "nls", formula = y ~ a * exp(b * x), se = FALSE, color = "red",
method.args = list(start = list(a = 5, b = 0))) + # Add a red
labs(title = "Exponetial model",
x = "X", y = "Y") + # Set title and axis labels
theme_minimal() 
Non-linear models
Modelsq quadratic regression model
Non Linear model using modelsq quadratic regression model not a good one
library(ggplot2)
y <- SalesCars$Cars_sold
x <- SalesCars$J_period
# Fit non-linear regression model
modelsq <- nls(y ~ a * x^2 + b * x + c, start = list(a = 1, b = 1, c = 1))
p <- ggplot(SalesCars, aes(x = J_period, y = Cars_sold)) +
geom_point() +
labs(x = "X", y = "Y")
# Add fitted line to the plot
p <- p +
stat_function(fun = function(x) predict(modelsq, newdata = data.frame(x = x)),
color = "red")
print(p)
Non-linear regression Output
print(summary(modelsq))##
## Formula: y ~ a * x^2 + b * x + c
##
## Parameters:
## Estimate Std. Error t value Pr(>|t|)
## a -0.0018521 0.0005004 -3.702 0.00303 **
## b 1.4094525 0.2306406 6.111 5.25e-05 ***
## c 63.8509693 19.6280432 3.253 0.00692 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 32.68 on 12 degrees of freedom
##
## Number of iterations to convergence: 1
## Achieved convergence tolerance: 3.252e-09